Suppose $(x_n), (y_n)$ are increasing and positive sequences such that $(y_n)$ is a Cauchy sequence and $$x_{n+1}-x_n<y_{n+1}-y_{n-1}$$ for all $n\in\mathbb N$. Then is $(x_n)$ also a Cauchy sequence?
I know that $x_{n+1}-x_n\to 0$ does not imply that $(x_n)$ is Cauchy but does the extra condition given above help? I tried to write
\begin{align} x_m-x_n &=x_{m}-x_{m-1}+x_{m-1}-x_{m-2}+\ldots+ x_{n+1}-x_n \\\\ &< y_m-y_{m-2}+y_{m-1}-y_{m-3}+\ldots+y_{n+1}-y_{n-1} \end{align}
but couldn't proceed.
The idea is to express $y_{n+1}-y_{n-1}$ as the sum of differences of adjacent sequence elements: $$ x_{n+1}-x_n<y_{n+1}-y_{n-1} = (y_{n+1}-y_n) + (y_n-y_{n-1}) $$ and therefore $$ 0 \le x_{n+p}-x_n = \sum_{k=0}^{p-1}(x_{k+1}-x_n) < \sum_{k=0}^{p-1}(y_{k+1}-y_n) + \sum_{k=0}^{p-1}(y_n-y_{k-1}) \\ = (y_{n+p}-y_n) + (y_{n+p-1}-y_{n-1}) $$ for all $n, p \ge 1$. Use this to show that if $(y_n)$ is a Cauchy sequence then $(x_n)$ is also a Cauchy sequence.
Note that the condition that $(y_n)$ is increasing it not needed for the proof.